and then from Theorem  that \(\delta _p(R)=\delta _p(L).\) 
But the only pole of \(L(\lambda ):=\lambda L_{1} + L_0\)  is the point at infinity and its polar degree is equal to \(\mbox{${\rm rank}(L_1)$}\)  [1]}. Therefore, the McMillan degree \(\delta _p(R)\)  of \(R(\lambda )\)  equals the rank of \(L_1\)  for any strongly minimal linearization of \(R(\lambda )\)  and no other pencils with the same zero structure and the same left and right minimal indices as \(R(\lambda )\)  can have a first order coefficient with smaller rank. Thus, strongly minimal linearizations are optimal in this sense.
